The equations $3x-2y+z=5,6x-4y+2z=10,9x-6y+3z=15$ have

$WhereA=\left[\begin{array}{ccc}3& -2& 1\\ 6& -4& 2\\ 9& -6& 3\end{array}\right]=3\left(-12+12\right)+2\left(28-18\right)+1\left(-36+36\right)$

$\left|A\right|=0$  and every determinant of $2\times 2$  is 0

So rank of $A\le 1<3$

$AD=\left[\begin{array}{cccc}3& -2& 1& 5\\ 6& -4& 2& 10\\ 9& -6& 3& 15\end{array}\right]$

$R_2\to R_2-2R_1$

$R_3\to R_3-3R_1$

$\left[\begin{array}{cccc}3& -2& 1& 5\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

$P\left(A\right)=P\left(AD\right)=1<3$

The rank of a matrix $\text{'}A\text{'}$  is less than 3 then given equations has

in finitely many solutions.